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Bar complex

Technique for constructing resolutions in homological algebra

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Source updated January 25, 2026

In mathematics, the bar complex, also called the bar resolution, bar construction, standard resolution, or standard complex, is a way of constructing resolutions in homological algebra. It was first introduced for the special case of algebras over a commutative ring by Samuel Eilenberg and Saunders Mac Lane, and Henri Cartan and Eilenberg and has since been generalized in many ways. The name "bar complex" comes from the fact that Eilenberg and Mac Lane used a vertical bar | as a shortened form of the tensor product $\otimes$ in their notation for the complex.

01Definition

Let $R$ be an algebra over a field $k$ , let $M_{1}$ be a right $R$ -module, and let $M_{2}$ be a left $R$ -module. Then, one can form the bar complex $\operatorname {Bar} _{R}(M_{1},M_{2})$ given by

$\cdots \rightarrow M_{1}\otimes _{k}R\otimes _{k}R\otimes _{k}M_{2}\rightarrow M_{1}\otimes _{k}R\otimes _{k}M_{2}\rightarrow M_{1}\otimes _{k}M_{2}\rightarrow 0\,,$

with the differential

${\begin{aligned}d(m_{1}\otimes r_{1}\otimes \cdots \otimes r_{n}\otimes m_{2})&=m_{1}r_{1}\otimes \cdots \otimes r_{n}\otimes m_{2}\\&+\sum _{i=1}^{n-1}(-1)^{i}m_{1}\otimes r_{1}\otimes \cdots \otimes r_{i}r_{i+1}\otimes \cdots \otimes r_{n}\otimes m_{2}+(-1)^{n}m_{1}\otimes r_{1}\otimes \cdots \otimes r_{n}m_{2}\end{aligned}}$

02Resolutions

The bar complex is useful because it provides a canonical way of producing (free) resolutions of modules over a ring. However, often these resolutions are very large, and can be prohibitively difficult to use for performing actual computations.

Free Resolution of a Module

Let $M$ be a left $R$ -module, with $R$ a unital $k$ -algebra. Then, the bar complex $\operatorname {Bar} _{R}(R,M)$ gives a resolution of $M$ by free left $R$ -modules. Explicitly, the complex is

$\cdots \rightarrow R\otimes _{k}R\otimes _{k}R\otimes _{k}M\rightarrow R\otimes _{k}R\otimes _{k}M\rightarrow R\otimes _{k}M\rightarrow 0\,,$

This complex is composed of free left $R$ -modules, since each subsequent term is obtained by taking the free left $R$ -module on the underlying vector space of the previous term.

To see that this gives a resolution of $M$ , consider the modified complex

$\cdots \rightarrow R\otimes _{k}R\otimes _{k}R\otimes _{k}M\rightarrow R\otimes _{k}R\otimes _{k}M\rightarrow R\otimes _{k}M\rightarrow M\rightarrow 0\,,$

Then, the above bar complex being a resolution of $M$ is equivalent to this extended complex having trivial homology. One can show this by constructing an explicit homotopy $h_{n}:R^{\otimes _{k}n}\otimes _{k}M\to R^{\otimes _{k}(n+1)}\otimes _{k}M$ between the identity and 0. This homotopy is given by

${\begin{aligned}h_{n}(r_{1}\otimes \cdots \otimes r_{n}\otimes m)&=\sum _{i=1}^{n-1}(-1)^{i+1}r_{1}\otimes \cdots \otimes r_{i-1}\otimes 1\otimes r_{i}\otimes \cdots \otimes r_{n}\otimes m\end{aligned}}$

One can similarly construct a resolution of a right $R$ -module $N$ by free right modules with the complex $\operatorname {Bar} _{R}(N,R)$ .

Notice that, in the case one wants to resolve $R$ as a module over itself, the above two complexes are the same, and actually give a resolution of $R$ by $R$ - $R$ -bimodules. This provides one with a slightly smaller resolution of $R$ by free $R$ - $R$ -bimodules than the naive option $\operatorname {Bar} _{R^{e}}(R^{e},M)$ . Here we are using the equivalence between $R$ - $R$ -bimodules and $R^{e}$ -modules, where $R^{e}=R\otimes R^{\operatorname {op} }$ , see bimodules for more details.

03The Normalized Bar Complex

The normalized (or reduced) standard complex replaces $A\otimes A\otimes \cdots \otimes A\otimes A$ with $A\otimes (A/K)\otimes \cdots \otimes (A/K)\otimes A$ .

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Sources and credits

This article is adapted from the Wikipedia article “Bar complex”, written by its contributors and licensed under CC BY-SA 4.0. Fathomly has changed the layout, removed citation markers, navigation and maintenance notices, and adjusted punctuation. This adapted version is shared under the same license. For references, see the original article.

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